Question

If $\cos x\dfrac{dy}{dx}-y\sin x=6x,\left( 0 < x < \dfrac{\pi }{2} \right)\text{ and }y\left( \dfrac{\pi }{3} \right)=0,\text{then }y\left( \dfrac{\pi }{6} \right)$ is equal to:

$\begin{aligned} & a.-\dfrac{{{\pi }^{2}}}{4\sqrt{3}} \\\ & b.-\dfrac{{{\pi }^{2}}}{2} \\\ & c.-\dfrac{{{\pi }^{2}}}{2\sqrt{3}} \\\ & d.\dfrac{{{\pi }^{2}}}{2\sqrt{3}} \\\ \end{aligned}$

Answer 1

In the question differential form of x and y functions are given. As we can see it is in the form of $\dfrac{dy}{dx}+y\left( P\left( x \right) \right)=Q\left( x \right)$, that is LDE (Linear Differential Equation). We solve it using the general method. Firstly, find the integrating factor, $IF={{e}^{\int{P\left( x \right)dx}}}$ and then using the formula $IFy=\int{IF.Q\left( x \right)}$ we will get a value for y.

Complete step-by-step answer:
In this question, we have been given the differential equation as $\cos x\dfrac{dy}{dx}-y\sin x=6x$. So, we will proceed by dividing the whole equation with cos x. So, we will get,
$\begin{aligned} & \dfrac{dy}{dx}-y\dfrac{\sin x}{\cos x}=\dfrac{6x}{\cos x} \\\ & \dfrac{dy}{dx}-y\tan x=\dfrac{6x}{\cos x}.........\left( 1 \right) \\\ \end{aligned}$
Now, before solving, we need to understand the concept of LDE, that is, linear differential equation. In mathematics, LDE is a differential equation that is defined by a linear polynomial in the unknown function and its derivative, that is an equation of the form, $\dfrac{dy}{dx}+y\left( P\left( x \right) \right)=Q\left( x \right)$. To solve this we need to take the following steps.
1: Find the integrating factor, $IF={{e}^{\int{P\left( x \right)dx}}}$
2: The formula of the solution of LDE is $IFy=\int{IF.Q\left( x \right)}$
So, here, we can see that,
$P\left( x \right)=-\tan x,Q\left( x \right)=\dfrac{6x}{\cos x}$
So, first we will find IF.

& IF={{e}^{\int{P\left( x \right)dx}}} \\\ & IF={{e}^{\int{-\tan x.dx}}} \\\ & IF={{e}^{-\int{\tan x.dx}}} \\\ & IF={{e}^{-\left( -\ln \left| \cos x \right| \right)}}=\cos x \\\ \end{aligned}$$ Now, putting this value in the formula, $IFy=\int{IF.Q\left( x \right)}$, we will get, $$\begin{aligned} & y\cos x=\int{\dfrac{6x}{\cos x}\times \cos xdx+C} \\\ & y\cos x=\int{6xdx+C} \\\ \end{aligned}$$ Now, we know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$, so we can write, $$\begin{aligned} & y\cos x=\dfrac{6{{x}^{2}}}{2}+C \\\ & y\cos x=3{{x}^{2}}+C \\\ \end{aligned}$$ Now, we have been given that at $x=\dfrac{\pi }{3},y=0$. So, to find C, we will substitute these values in the above equation. So, we get, $\begin{aligned} & 0=3\times {{\left( \dfrac{\pi }{3} \right)}^{2}}+C \\\ & 0=\dfrac{{{\pi }^{2}}}{3}+C \\\ & C=-\dfrac{{{\pi }^{2}}}{3} \\\ \end{aligned}$ As $C=-\dfrac{{{\pi }^{2}}}{3}$, we finally get our solution as, $y\cos x=3{{x}^{2}}-\dfrac{{{\pi }^{2}}}{3}$ Now, according to the question, we need to find $y\left( \dfrac{\pi }{6} \right)$. So, we will put $x=\dfrac{\pi }{6}$ in $y\cos x=3{{x}^{2}}-\dfrac{{{\pi }^{2}}}{3}$, so we will get, $y\cos \left( \dfrac{\pi }{6} \right)=3{{\left( \dfrac{\pi }{6} \right)}^{2}}-\dfrac{{{\pi }^{2}}}{3}$ We know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$, so we get, $\begin{aligned} & y\times \dfrac{\sqrt{3}}{2}=3\times {{\dfrac{\pi }{36}}^{2}}-\dfrac{{{\pi }^{2}}}{3} \\\ & \dfrac{y\sqrt{3}}{2}=\dfrac{{{\pi }^{2}}}{12}-\dfrac{{{\pi }^{2}}}{3} \\\ & \dfrac{y\sqrt{3}}{2}=\dfrac{{{\pi }^{2}}-4{{\pi }^{2}}}{12} \\\ & y=\dfrac{-3{{\pi }^{2}}}{12}\times \dfrac{2}{\sqrt{3}} \\\ & y=-\dfrac{{{\pi }^{2}}}{2\sqrt{3}} \\\ \end{aligned}$ Therefore, we get $y\left( \dfrac{\pi }{6} \right)$ as $-\dfrac{{{\pi }^{2}}}{2\sqrt{3}}$. **So, the correct answer is “Option C”.** **Note:** In case of questions related to differential equations of LDE, it is by solved using the same steps. We need to keep in mind that while solving IF, that is ${{e}^{\int{P\left( x \right)dx}}}$, after finding $$\int{P\left( x \right)dx}$$, there is no addition of the integration constant. Also, the value of $0